7.2. CALCULUS OF VARIATIONS c 2006 Gilbert Strang 7.2 Calculus of Variations One theme of this book is the relation of equations to minimum principles. To minimize P is to solve P 0 = 0. There may be more to it, but that is the main 1 T T point. For a quadratic P (u) = 2 u Ku u f, there is no difficulty in reaching P = Ku f = 0. The matrix K is symmetric− positive definite at a minimum. 0 − In a continuous problem, the \derivative" of P is not so easy to find. The unknown u(x) is a function, and P (u) is usually an integral. Its derivative δP/δu is called the first variation. The \Euler-Lagrange equation" δP/δu = 0 has a weak form and a strong form. For an elastic bar, P is the integral of 1 c(u (x))2 f(x)u(x). 2 0 − The equation δP/δu = 0 is linear and the problem will have boundary conditions: Weak form cu v dx = fv dx for every v Strong form (cu ) = f(x): 0 0 − 0 0 Z Z Our goal in this section is to get beyond this first example of δP/δu. The basic idea should be simple and it is: Perturb u(x) by a test function v(x). Comparing P (u) with P (u + v), the linear term in the difference yields δP/δu. This linear term must be zero for every admissible v (weak form). This program carries ordinary calculus into the calculus of variations. We do it in several steps: 1. One-dimensional problems P (u) = F (u; u 0) dx, not necessarily quadratic R 2. Constraints, not necessarily linear, with their Lagrange multipliers 3. Two-dimensional problems P (u) = F (u; ux; uy) dx dy RR 4. Time-dependent equations in which u 0 = du=dt. At each step the examples will be as familiar (and famous) as possible. In two dimensions that means Laplace's equation, and minimal surfaces in the nonlinear case. In time-dependent problems it means Newton's Laws, and relativity in the nonlinear case. In one dimension we rediscover the straight line and the circle. This section is also the opening to control theory|the modern form of the calculus of variations. Its constraints are differential equations, and Pontryagin's maximum principle yields solutions. That is a whole world of good mathematics. Remark To go from the strong form to the weak form, multiply by v and integrate. For matrices the strong form is ATCAu = f. The weak form is vTATCAu = vTf for all v. For functions with Au = u 0, this exactly matches cu 0v 0 dx = fv dx above. R R c 2006 Gilbert Strang One-dimensional Problems The basic problem is to minimize P (u) with a boundary condition at each end: 1 P (u) = F (u; u 0) dx with u(0) = a and u(1) = b : Z0 The best u defeats every other candidate u+v that satisfies these boundary conditions. Then (u + v)(0) = a and (u + v)(1) = b require that v(0) = v(1) = 0. When v and v 0 are small the correction terms come from @F=@u and @F=@u 0. They don't involve v2: @F @F Inside the integral F (u + v; u 0 + v 0) = F (u; u 0) + v + v 0 + @u @u 0 · · · 1 @F @F After integrating P (u + v) = P (u) + v + v dx + @u 0 @u · · · Z0 0 That integrated term is the “first variation". We have already reached δP/δu: δP 1 @F @F Weak form = v + v dx = 0 for every v : (1) δu @u 0 @u Z0 0 This is the equation for u. The derivative of P in each direction v must be zero. Otherwise we can make δP/δu negative, which would mean P (u + v) < P (u): bad. The strong form looks for a single derivative which|if it is zero|makes all these directional derivatives zero. It comes from integrating δP/δu by parts: 1 @F d @F @F 1 Weak form / by parts v v dx + v = 0 : @u − dx @u @u Z0 0 0 0 The boundary term vanishes because v(0) = v(1) = 0. To guarantee zero for every v(x) in the integral, the function multiplying v must be zero (strong form): @F d @F Euler-Lagrange equation for u = 0 : (2) @u − dx @u 0 Example Find the shortest path u(x) between two points (0; a) and (1; b). 2 2 2 By Pythagoras, (dx) + (du) is a short step on the path. So P (u 0) = 1 + (u 0) dx is the length of the path between the points. This square root F (u ) depends only on u p 0 R p 0 and @F=@u = 0. The derivative @F=@u 0 brings the square root into the denominator: @F 1 u Weak form = v 0 dx = 0 for every v with v(0) = v(1) = 0 . (3) 0 2 @u 0 1 + (u ) Z 0 p If the quantity multiplying v 0 is a constant, then (3) is satisfied. The integral is certain to be zero because v(0) = v(1) = 0. The strong form forces @F=@u 0 to be 7.2. CALCULUS OF VARIATIONS c 2006 Gilbert Strang constant: the Euler-Lagrange equation (2) is d @F d u u = 0 = 0 or 0 = c : (4) −dx @u −dx 2 2 0 1 + (u 0) 1 + (u 0) 0 That integration is always possiblep when F dependsp only on u (@F=@u = 0). It leaves the equation @F=@u 0 = c. Squaring both sides, u is seen to be linear: 2 2 c c (u 0) = c (1 + (u 0)2) and u 0 = and u = x + d : (5) p1 c2 p1 c2 − − PSfrag replacementsThe constants c and d are chosen to match u(0) = a and u(1) = b. The shortest curve connecting two points is a straight line. No surprise! The length P (u) is a minimum, not a maximum or a saddle point, because the second derivative F 00 is positive. perturbed u = optimal arc u + v b b 1 1 −m a u = optimal line a −m x x 0 v 1 0 1 typical perturbation v(x) area below arc is A Figure 7.5: Shortest paths from a to b: straight line and circular arc (constrained). Constrained Problems Suppose we cannot go in a straight line because of a constraint. When the constraint is u(x) dx = A, we look for the shortest curve that has area A below it: R 1 1 2 Minimize P (u) = 1 + (u 0) dx with u(0) = a; u(1) = b; u(x) dx = A : 0 0 Z p Z The area constraint should be built into P by a Lagrange multiplier|here called m. The multiplier is a number and not a function, because there is one overall constraint rather than a constraint at every point. The LagrangianL builds in u dx = A: Lagrangian L(u; m) = P + (multiplier)(constraint) = (F + mu)Rdx mA : − R The Euler-Lagrange equation δL/δu = 0 is exactly like δP/δu = 0 in (2): @(F + mu) d @(F + mu) d u = m 0 = 0 : (6) @u − dx @u − dx 2 0 1 + (u 0) p c 2006 Gilbert Strang Again this equation is favorable enough to be integrated: u 0 mx c mx = c which gives u 0 = − : − 1 + (u )2 1 (mx c)2 0 − − After one more integrationp we reach the equation of a circlep in the x-u plane: 1 u(x) = − 1 (mx c)2 + d and (mx c)2 + (mu d)2 = 1 : (7) m − − − − The shortest pathpis a circular arc! It goes high enough to enclose area A. The three numbers m; c; d are determined by the conditions u(0) = a; u(1) = b, and u dx = A. The arc is drawn in Figure 7.5 (and m is negative). R We now summarize the one-dimensional case, allowing F to depend also on u 00. That introduces v 00 into the weak form and needs two integrations by parts to reach the Euler-Lagrange equation. When F involves a varying coefficient c(x), the form of the equation does not change, because it is u and not x that is perturbed. 1 0 00 The first variation of P (u) = 0 F (u; u ; u ; x) dx is zero at a minimum: R Weak δP 1 @F @F @F = v + v + v dx = 0 for all v: form δu @u 0 @u 00 @u Z0 0 00 The Euler-Lagrange equation from integration by parts determines u(x): Strong @F d @F d2 @F + = 0 : form @u − dx @u dx2 @u 0 00 Constraints on u bring Lagrange multipliers and saddle points of L. Applications are everywhere, and we mention one (of many) in sports. What angle is optimal in shooting a basketball? The force of the shot depends on the launch angle|line drives or sky hooks need the most push. The force is minimized at 45◦ if the ball leaves your hand ten feet up; for shorter people the angle is about 50◦. What is interesting is that the same angle solves a second optimization problem: to have the largest margin of error and still go through the hoop. The condition is P 0 = 0 in basketball (one shot) and δP/δu = 0 in track|where the strategy to minimize the time P (u) has been analyzed for every distance.